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3 years ago

Need Help With Physical Ed

Physics

2 answers:

Dimas [21]3 years ago

8 0

which of the following is not a barrier to physical activity it is fear of injury I think.

mote1985 [20]3 years ago

5 0

I believe it’s fear of injury

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An inner city revitalization zone is a rectangle that is twice as long as it is wide. The width of the region is growing at a ra

lukranit [14]

Answer:

$\frac{dA}{dt} = 28800 \ m^2/year$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Geometry

Area of a Rectangle: A = lw

Algebra I

Exponential Property: $w^n \cdot w^m = w^{n + m}$

Calculus

Derivatives

Differentiating with respect to time

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Explanation:

Step 1: Define

Area is A = lw

2w = l

w = 300 m

$\frac{dw}{dt} = 24 \ m/year$

Step 2: Rewrite Equation

Substitute in l: A = (2w)w
Multiply: A = 2w²

Step 3: Differentiate

Differentiate the new area formula with respect to time.

Differentiate [Basic Power Rule]: $\frac{dA}{dt} = 2 \cdot 2w^{2-1}\frac{dw}{dt}$
Simplify: $\frac{dA}{dt} = 4w\frac{dw}{dt}$

Step 4: Find Rate

Use defined variables

Substitute: $\frac{dA}{dt} = 4(300 \ m)(24 \ m/year)$
Multiply: $\frac{dA}{dt} = (1200 \ m)(24 \ m/year)$
Multiply: $\frac{dA}{dt} = 28800 \ m^2/year$

3 0

3 years ago

Read 2 more answers

Key Stage 3 Science - Physics

stira [4]

Answer:

F = 3750 N

Explanation:

Given that,

Pressure, P = 150 Pa

Area, a = 25m²

We need to find the force applied. We know that, pressure is equal to the force acting per unit area. It can be given by :

$P=\dfrac{F}{A}\\\\F=P\times A\\\\F=150\ Pa\times 25\ m^2\\\\F=3750\ N$

So, the required force is 3750 N.

5 0

3 years ago

Un vagón de carga de 5000-kg se mueve a 2m/s e impacta a un vagón de 10000-kg que se encuentra en reposo. Después de la colisión

Mariulka [41]

Answer:

0.67 m/s

Explanation:

Mass of car 1, m₁ = 5000 kg

Mass of car 2, m₂ = 10,000 kg

Initial speed of car 1, u₁ = 2 m/s

Final speed of car 2, u₂ = 0 (at rest)

We need to find the final velocity of both cars when inelastic collision occurs. The momentum will remain conserved in case of inelastic collision. Using the conservation of momentum. Let V is the final speed.

$m_1u_1+m_2u_2=(m_1+m_2)V\\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{m_1u_1}{(m_1+m_2)}\\\\V=\dfrac{5000\times 2}{(5000+10000)}\\\\V=0.67\ m/s$

So, after the inelastic collision, they will move with a speed of 0.67 m/s.

6 0

3 years ago

I'll give brainliest can someone help me with 2.4.1-2.4.3?with explanation.

KIM [24]

Answer:

220÷4=550 per hour

2.4.3 550×4=2200

hope it's right

8 0

3 years ago

HELP PLZZZZZZ

ValentinkaMS [17]

Hi there!

We know that:

U (Potential energy) = mgh

We are given the potential energy, so we can rearrange to solve for h (height):

U/mg = h

g = 9.81 m/s²

m = 30 g ⇒ 0.03 kg

0.062/(0.03 · 9.81) = 0.211 m

8 0

3 years ago